The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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What kind of Choice am I making in this argument?

I have an argument that's supposed to imply Choice, but I'm afraid it may be using some choice. If it does, how much choice? This is the part of the argument that might use some Choice. I marked the ...
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Infinite Cardinal Addition Without the Axiom of Choice

In the book 'Introduction to Set Theory' by Hrbacek and Jech, cardinal addition is defined as $$\sum_{i \in I}{\kappa_i}=\left|\bigcup_{i \in I}{A_i} \right|$$ where $|A_i|=k_i$ for all $i \in I$ ...
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Cardinality of the union of two infinite set

Suppose that $A$ and $B$ are two infinite sets and $|A|<|B|$. The question is that how to prove that $|A∪B|=|B|$. The proof is related to the Axiom of Choice.
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Existence of non-trivial linear functional on any vector space

For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , ...
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$\mathbf{Q}$ basis of $\mathbf{R}$.

Could someone give me an explicit basis of $\mathbf{R}$ as a vector space over $\mathbf{Q}$? I no some linearly independent subset, namely $1,e,e^2,\ldots$ but this seems to be a deep result already ...
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Does the existence of a recursive sequence that involves an arbitrary choice at every step require the axiom of choice?

Say I want to define a sequence $(x_n)$ recursively, and at each step I make an arbitrary selection for $x_{n+1}$ out of some nonempty pool of acceptable candidates dependent on $x_n$. Does the ...
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65 views

How big can the continuum be without choice?

I've heard an argument before (although I can't remember where) that the continuum hypothesis is false, since the powerset operation is a something much more 'powerful' than the mere cardinal ...
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Over ZF, does “every Hilbert space have a basis” imply AC?

I know there is a similar result due to Blass [1] that over ZF, "every vector space has a (Hamel) basis" implies AC. Looking around, however, I can't find any results on the question for Hilbert ...
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Existence of a basis in constructive vector spaces

As I was trying to review forgotten knowledge on Vector Spaces in wikipedia, I read that the existence of a basis follows from Zorn lemma, hence equivalently from the axiom of choice. Actually, the ...
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44 views

Dedekind finite set and a special well ordered set

In ZFC, Dedekind finite set and finite set are same things. So I have a set say A(which is equal to N in ZFC) all Dedekind finite set are equivalent to proper subsets of A and A is well ordered set. ...
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Basis in topology and AC

Let S ⊂ ℘(X). Let T be the coarsest topology on X which contains S. Then we call T the topology generated by S. Let S ⊂ ℘(X). Then we can easily prove using (generalised distributive law) the ...
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Countable union of countable sets is countable and AC(or axiom of countable choice)

It is written in tag page of Axiom of choice in MSE that Countable union of countable sets is countable is a theorem which follows from AC. Do we really need AC to prove this. Please see following ...
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What is needed to make Euclidean spaces isomorphic as groups?

Consider the abelian groups $G_n=(\mathbb R^n,+)$ for $n\geq1$. Claim: For any $n$ and $m$ the groups $G_n$ and $G_m$ are isomorphic. This claim is true if one assumes the axiom of choice, and I ...
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45 views

Existence of non measurable set and ZF theory?

Does statement: 'Existence of non measurable set' consistent with ZF theory. or if I throw Axiom of choice from ZFC theory. Can I prove or disprove existence of a countably additive measure function ...
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Does there exist a connected 2-regular uncountable graph, or an uncountable path?

Does there exist a connected 2-regular uncountable graph? Can I use the axiom of choice to construct an uncountable path of elements from the reals? The question arose when reading this: Also, ...
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How does one prove $ZF\vDash MC\Rightarrow AC$?

This is somewhat adressed to Andreas Blass, whose papers I have read, in particular I make reference to an old paper of his »Existence of Basis implies the Axiom of Choice« (84). Anyone who happens to ...
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First Order Model Existence without Weak Konigs Lemma or Choice

In studying Godel's Completeness Theorem and its various related formulations like the Model Existence Theorem and the Lowenheim-Skolem Theorem there is one rather subtle point that I have not yet ...
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78 views

Infinite Set has greater or equal cardinality that of N

For any infinite set, we can find a 1-1 function (not necessary onto) from N (set of natural no.) to that set. The proof of this theorem I know using axiom of choice. Can we prove it without using ...
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42 views

Choice function to obtain all elements of an infinite set

I'm following a proof of "Axiom of Choice implies Well Ordering Theorem" (from a not-well-known book we use for class), and the author uses this procedure: First, he takes any non-empty set $X$ and ...
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1answer
66 views

Does this argument rely on countable choice?

Consider the following Theorem: Any algebraic field extension $K|F$ of infinite degree contains finite subextensions of arbitrarily high degree. Proof: We'll prove that, for any n, there's a ...
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Proof of one instance of the Axiom of Choice from another.

I'm trying to show (i) $\implies$ (ii): (i) For any relation $R$, there exists a function $H\subseteq R$, with $\newcommand{\dom}{\mathrm{dom} \ } \dom H = \dom R$. (ii) For any set $I$ and any ...
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Proving equivalence of Axiom of Choice

I am working on the following question concerning the axiom of choice and one of its many equivalences. Advice as to whether I am on the right track would be appreciated. As a preface, I have looked ...
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divisible subgroup without axiom of choice

the theorem asserting that the divisible subgroup of an Abelian group is a direct summand depends on Zorn's lemma. in ZF without AC is there a construction which yields a model of an Abelian group ...
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Absolute coequalizers in $\mathbf {Set} $

Let $ A $ be a set and let $ R\subseteq A\times A $ be an equivalence relation on $ A $. Denote by $ p, q $ the projections $ R\longrightarrow A $ on the first and second factor, respectively. The ...
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Miller's Construction, Partition Principle and Failure of Axiom of Choice

Partition Principle ($PP$) is the following statement: For all sets $a$, $b$ there is an injection $f:a\rightarrow b$ iff there is a surjection $g:b\rightarrow a$ It is known that $ZF\vdash ...
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Is this an equivalent form of Axiom of Choice? [duplicate]

It is known that Axiom of Choice implies the following statement: For each two sets $A$ and $B$, there is a one to one function from $A$ to $B$ iff there is a function from $B$ onto $A$ Is above ...
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Why is the powerset axiom more acceptable than the axiom of choice?

The key step in Zermelo's proof of the well ordering theorem is to use $\text{AC}$ to simultaneously choose the next elelment for all possible partial chains in prospective well orderings, but that ...
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41 views

Existence of a Set Function Axiom of Choice

I have the following problem. Let $A$ be a set and $B\neq\emptyset$ be a proper subset. Prove the existence of a function $f:A\to A$ such that $f\circ f=f$ and $\text{im}~f=B$. In the case where $A$ ...
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73 views

Does the Cartesian product of an infinite family have all the elements we expect?

Given the axiom of choice, we know that the Cartesian product of an infinite family of non-empty sets is non-empty. However, this doesn't tell us whether the Cartesian product contains every element ...
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Does the converse of Tychonoff's theorem hinge on the axiom of choice?

Tychonoff's theorem:$\phantom{---}$ If $A$ is a non-empty index set and $X_{\alpha}$ is a non-empty compact topological space for every $\alpha\in A$, then $X\equiv\times_{\alpha\in A} X_{\alpha}$ is ...
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Disjoint $AC$ equivalent to $AC$

I want to show that these two forms of $AC$ are equivalent: $(1)$ For each collection of nonempty sets $X$ there is a choice function. $(2)$ For each collection of pairwise disjoint nonempty sets $X$ ...
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Axiom of choice variants [duplicate]

Is there any good book where the equivalence of AC to the statement "Any surjection has a right inverse" is proved (and maybe other equivalences)? I could do $AC \Rightarrow$ "Any surjection has a ...
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Formal approach to (countable) prisoners and hats problem.

I've found this nice puzzle about AC (I'm referring to the countable infinite case, with two colors). The puzzle has been discussed before on math.SE, but I can't find any description of what is ...
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Characterization of dense open subsets of the real numbers

Does the complement of every dense open subset of the real numbers have Lebesgue measure $0$? This is certainly not a characterization of dense open subsets of reals, since the complement of the ...
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Proof of “Every vector space has a basis $\implies$ AC” without mentioning von Neumann hierarchy

I am writing a short (30-50 pages) report on AC for an exam. I really would like to include the proof that "Every vector space has a basis $\implies$ AC". Actually, every proof I could find proves ...
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Is “PA has no non-standard models” consistent with ZF?

I have seen several proofs that there exist nonstandard models of arithmetic, but they all seem to rely on the compactness theorem, which is not implied by ZF. So are there any proofs in ZF that ...
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Writing the ZF formula for the choice function (given Well-ordering)

If every set has a well order, then the axiom of choice follows: Given a well order on $\bigcup_{i \in I} A_i$ we define the choice function in this way $f(i) =$ "the first element of $\bigcup_{i \in ...
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What is the definition of the Feferman-Levy model?

Any (reference to) definition of Feferman-Levy model in set theory? I cannot find any... Though I know what is Levy collapse.
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Proof of: $X$ is finite $\iff X$ is Tarski-finite

I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876). In the fourth chapter, he deals with different definitions of finite set. Here is the classical one: ...
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Examples of figures whose area depend on axiom of choice

Many times I have heard that there are some 'figures' whose area is not fixed and they depend on axiom of choice. For example, In this answer. In this lecture about probabilty. I did not know ...
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An explicit construction for an everywhere discontinuous real function with $F((a+b)/2)\leq(F(a)+F(b))/2$?

I would like to know an explicit method on constructing an everywhere discontinuous real function $F$ with the property: $$F((a+b)/2)\leq(F(a)+F(b))/2.$$ There is a non-constructive example (with the ...
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How do we get a simplicial homology functor?

The $n$-th simplicial homology group $H_n(A)$ of an abstract simplicial complex $A$ depends on the choice of an orientation for $A$ (but for different orientations, the homology groups are ...
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Is the Axiom of Choice necessary to prove $\mathbb R \approx \mathcal P(\omega)$?

I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876), and I'm getting quite confused about cardinal arithmetic without AC. Here (Which sets are well-orderable ...
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Why do we need the axiom of choice in showing the non-emptiness of an infinite Cartesian product

Given $I$ a set of indexes and $X_i$ a set of topological spaces, define The Cartesian product: $\prod_{i \in I}X_i = \{ f:I \rightarrow \bigcup X_i | f(i) \in X_i \}$ I have read that we need ...
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infinite Cartesian products

Quote from Wikipedia, infinite set: The Cartesian product of an infinite number of sets each containing at least two elements is either empty or infinite. I know that this is the regime where we ...
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Is the Axiom of Choice restricted to $\Bbb R$ independent of ZF? [duplicate]

This is really a yes-no question, and I am hundred percent certain that the answer is "yes". I simply have not found it written directly anywhere. True, the Axiom of Choice is independent of ZF if we ...
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Could induction give us an infinite sequence of sets $X_1 \subseteq X_2 \subseteq \cdots$ or do we need the axiom of choice?

Suppose at each step $n$ of a proof by induction, we have a set $X_n$. Also, $X_{n} \subseteq X_{n+1}$. Is induction enough to give us $\bigcup_{n \in \mathbb{N}} X_n$? Or rather, when is it enough? ...
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Partitions of the Cantor space into parities

Call a partition of $2^\mathbb{N} = A\cup B$ a parity partition if, for any $n\in\mathbb{N}$, flipping the $n$th bit of any element of $A$ results in an element of $B$, and vice-versa. Given a choice ...
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Can we prove without the axiom of choice that subspaces of finite dimensional vector spaces are finite dimensional as well?

Let $E$ be a vector space and $F$ a subspace of $E$. Show that if $E$ is finite dimensional then $F$ is finite dimensional too. It's easy to prove by contradiction by taking a family linearly ...
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Bernstein sets, Well-Ordering theorem vs Axiom of Choice

In the construction of Bernstein sets (see here), is it necessary to use the well-ordering theorem? Why can't you just use the Axiom of Choice to pick two points?